Problem: What is the value of $\dfrac{d}{dx}\tan(x)$ at $x=\pi$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $\dfrac12$ (Choice C) C $0$ (Choice D) D $1$
Let's first find $\dfrac{d}{dx}\tan(x)$. Then, we can evaluate it at $x=\pi$. Recall that the derivative of $\tan(x)$ is $\dfrac{1}{\cos^2(x)}$, or $\sec^2(x)$. Put another way, $\dfrac{d}{dx}[\tan(x)]=\dfrac{1}{\cos^2(x)}=\sec^2(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\pi}$ : $\begin{aligned} &\phantom{=}\dfrac{1}{\cos^2\left({\pi}\right)} \\\\ &=\dfrac{1}{(-1)^2} \\\\ &=1 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\tan(x)$ at $x=\pi$ is $1$.